3D Viewing & Clipping
Where
Where do
do geometries
geometries come
come from?
from?
Pin-hole
camera
Pin-hole camera
Perspective
Perspective projection
projection
Viewing
Viewing transformation
transformation
Clipping
Clipping lines
lines &
& polygons
polygons
Angel Chapter 5
Getting Geometry on the Screen
Given geometry in the world coordinate system,
how do we get it to the display?
•
•
•
•
•
Transform to camera coordinate system
Transform (warp) into canonical view volume
Clip
Project to display coordinates
(Rasterize)
1
Vertex Transformation Pipeline
Vertex
ModelView
matrix
Eye coordinates
Projection
matrix and
perspective
division
Image plane coordinates
Viewport
transformation
Window coordinates
Vertex Transformation Pipeline
Vertex
glMatrixMode(GL_MODELVIEW)
ModelView
matrix
Eye coordinates
glMatrixMode(GL_PROJECTION)
Projection
matrix and
perspective
division
Image plane coordinates
glViewport(…)
Viewport
transformation
Window coordinates
2
OpenGL Transformation Overview
glMatrixMode(GL_MODELVIEW)
gluLookAt(…)
glMatrixMode(GL_PROJECTION)
glFrustrum(…)
gluPerspective(…)
glOrtho(…)
glViewport(x,y,width,height)
Viewing and Projection
• Our eyes collapse 3-D world to 2-D retinal image
(brain then has to reconstruct 3D)
• In CG, this process occurs by projection
• Projection has two parts:
– Viewing transformations: camera position and direction
– Perspective/orthographic transformation: reduces 3-D
to 2-D
• Use homogeneous transformations
• As you learned in Assignment 1, camera can be
animated by changing these transformations—
the root of the hierarchy
3
Pinhole Optics
• Stand at point P, and look through the hole - anything within the
cone is visible, and nothing else is
• Reduce the hole to a point - the cone becomes a ray
• Pin hole is the focal point, eye point or center of projection.
F
P
Perspective Projection of a Point
Image
W
F
I
World
• View plane or image plane - a plane behind the
pinhole on which the image is formed
–point I sees anything on the line (ray) through the
pinhole F
–a point W projects along the ray through F to appear at
I (intersection of WF with image plane)
4
Image Formation
Image
F
World
• Projecting a shape
– project each point onto the image
plane
– lines are projected by projecting end
points only
Image
F
W
I
World
Note:
Note: Since
Since we
we don’t
don’t want
want the
the
image
image to
to be
be inverted,
inverted, from
from now
now on
on
we’ll
we’ll put
put FF behind
behind the
the image
image plane.
plane.
Orthographic Projection
• when the focal point is at infinity the rays are parallel
and orthogonal to the image plane
• good model for telephoto lens. No perspective effects.
• when xy-plane is the image plane (x,y,z) -> (x,y,0)
front orthographic view
World
Image
F
5
A Simple Perspective Camera
• Canonical case:
–camera looks along the z-axis
–focal point is the origin
–image plane is parallel to the xy-plane at distance d
– (We call d the focal length, mainly for historical reasons)
y
Image
Plane
x
F=[0,0,0]
z
[0,0,d]
Similar Triangles
Y
[Y, Z]
[(d/Z)Y, d]
Z
[0, 0]
[0, d]
• Diagram shows y-coordinate, x-coordinate is similar
• Using similar triangles
– point [x,y,z] projects to [(d/z)x, (d/z)y, d]
6
A Perspective Projection Matrix
•Projection using homogeneous coordinates:
– transform [x, y, z] to [(d/z)x, (d/z)y, d]
ªd
«
«0
«0
«
«
¬0
0
d
0
0
0
0
d
1
0ºªxº
»« »
0»«y»
ªd
>dx dy dz z @ Ÿ « x
»
«
»
¬z
0»«z »
0»
¼
¬1»
¼«
th
º
d
y d»
¼
z
Divide by 4 coordinate
(the “w” coordinate)
• 2-D image point:
– discard third coordinate
– apply viewport transformation to obtain physical pixel coordinates
Wait, there’s more!
We have just seen how to project the entire world
onto the image plane..
How can we limit the portions of the scene that
are considered?
7
The View Volume
• Pyramid in space defined by focal point and window in
the image plane (assume window mapped to viewport)
• Defines visible region of space
• Pyramid edges are clipping planes
• Frustum = truncated pyramid with near and far clipping
planes
– Why near plane? Prevent points behind the camera being seen
– Why far plane? Allows z to be scaled to a limited fixed-point
value (z-buffering)
But wait...
• What if we want the camera somewhere other
than the canonical location?
• Alternative #1: derive a general projection
matrix. (hard)
• Alternative #2: transform the world so that the
camera is in canonical position and orientation
(much simpler)
• These transformations are viewing
transformations
8
Camera Control Values
• All we need is a single translation and angle-axis
rotation (orientation), but...
• Good animation requires good camera control--we need
better control knobs
• Translation knob - move to the lookfrom point
• Orientation can be specified in several ways:
– specify camera rotations
– specify a lookat point (solve for camera rotations)
A Popular View Specification Approach
• Focal length, image size/shape and clipping planes are in the
perspective transformation
• In addition:
– lookfrom:
– lookat:
where the focal point (camera) is
the world point to be centered in the image
• Also specify camera orientation about the lookat-lookfrom
axis
9
Implementation
Implementing the lookat/lookfrom/vup viewing scheme
(1) Translate by -lookfrom, bring focal point to origin
(2) Rotate lookat-lookfrom to the z-axis with matrix R:
» v = (lookat-lookfrom) (normalized) and z = [0,0,1]
» rotation axis:
a = (vxz)/|vxz|
» rotation angle:
cosT = v•z and sinT = |vxz|
glRotate(q, ax, ay, az)
(3) Rotate about z-axis to get vup parallel to the y-axis
The Whole Picture
LOOKFROM:
LOOKAT:
Where the camera is
A point that should be centered
in the image
VUP:
A vector that will be pointing
straight up in the image
FOV:
Field-of-view angle.
d:
focal length
WORLD COORDINATES
10
y
It's not so complicated…
x
vup
y
z
x
lookfrom
y
x
z
START HERE
z
lookat
Translate LOOKFROM
to the origin
Rotate the view vector
(lookat -lookfrom) onto
the z-axis.
y
x
Multiply by the projection matrix
and everything will be in the
canonical camera position
z
Rotate about z to bring vup to y-axis
One and Two-Point Perspective?
11
Clipping
• There is something missing between projection and
viewing...
• Before projecting, we need to eliminate the portion of
scene that is outside the viewing frustum
y
clipped line
x
z
image plane
near
far
•Need to clip objects to the frustum (truncated pyramid)
•Now in a canonical position but it still seems kind of tricky...
Normalizing the Viewing Frustum
• Solution: transform frustum to a cube before clipping
y
y
clipped line
clipped line
x
x
1
1
near
z
far
0
1
z
image plane
near
far
• Converts perspective frustum to orthographic frustum
• This is yet another homogeneous transform!
12
Clipping to a Cube
• Determine which parts of the scene lie within
cube
• We will consider the 2D version: clip to
rectangle
• This has its own uses (viewport clipping)
• Two approaches:
–clip during scan conversion (rasterization) - check per
pixel or end-point
–clip before scan conversion
• We will cover
– clip to rectangular viewport before scan conversion
Line Clipping
• Modify endpoints of lines to lie in rectangle
• How to define “interior” of rectangle?
• Convenient definition: intersection of 4 half-planes
–Nice way to decompose the problem
–Generalizes easily to 3D (intersection of 6 half-planes)
ymax
=
interior
ymin
xmin
ˆ
y < ymax
x > xmin
xmax
13
y > ymin
x < xmax
Line Clipping
• Modify end points of lines to lie in rectangle
• Method:
–Is end-point inside the clip region? - half-plane tests
–If outside, calculate intersection between the line and
the clipping rectangle and make this the new end
point
•
•
•
Both endpoints inside: trivial
accept
One inside: find intersection
and clip
Both outside: either clip or
reject (tricky case)
Cohen-Sutherland Algorithm
• Uses outcodes to encode the half-plane tests results
1001
1000
1010
ymax
0001
0000
0010
bit 1: y>ymax
bit 2: y<ymin
bit 3: x>xmax
bit 4: x<xmin
ymin
0101
0100
xmin
0110
xmax
• Rules:
– Trivial accept: outcode(end1) and outcode(end2) both zero
– Trivial reject: outcode(end1) & (bitwise and) outcode(end2)
nonzero
– Else subdivide
14
Cohen-Sutherland Algorithm: Subdivision
• If neither trivial accept nor reject:
–Pick an outside endpoint (with nonzero outcode)
–Pick an edge that is crossed (nonzero bit of outcode)
–Find line's intersection with that edge
–Replace outside endpoint with intersection point
–Repeat until trivial accept or reject
1001
1000
1010
0000
0010
ymax
0001
bit 1: y>ymax
bit 2: y<ymin
bit 3: x>xmax
bit 4: x<xmin
ymin
0101
0100
xmin
0110
xmax
Polygon Clipping
Convert a polygon into one or more polygons that
form the intersection of the original with the clip
window
15
Sutherland-Hodgman
Polygon Clipping Algorithm
• Subproblem:
–clip a polygon (vertex list) against a single clip plane
–output the vertex list(s) for the resulting clipped
polygon(s)
• Clip against all four planes
–generalizes to 3D (6 planes)
–generalizes to any convex clip polygon/polyhedron
Sutherland-Hodgman
Polygon Clipping Algorithm (Cont.)
To clip vertex list against one half-plane:
• if first vertex is inside - output it
• loop through list testing inside/outside transition - output
depends on transition:
> in-to-in:
> out-to-out:
> in-to-out:
> out-to-in:
output vertex
no output
output intersection
output intersection and vertex
16
Cleaning Up
• Post-processing is required when clipping creates
multiple polygons
• As external vertices are clipped away, one is left with
edges running along the boundary of the clip region.
• Sometimes those edges dead-end, hitting a vertex on
the boundary and doubling back
–Need to prune back those edges
• Sometimes the edges form infinitely-thin bridges
between polygons
–Need to cut those polygons apart
Ivan Sutherland
17
Beyond Linear Perspective...
Announcements
Written assignment #1 due next Thursday
18
Virtual Trackballs
• Imagine world contained in crystal ball, rotates about
center
• Spin the ball (and the world) with the mouse
• Given old and new mouse positions
– project screen points onto the sphere surface
– rotation axis is normal to plane of points and sphere center
– angle is the angle between the radii
• There are other methods to map screen coordinates to
rotations
19
Problems with Pinholes
• Correct optics requires infinitely small pinhole
– No light gets through
– Diffraction
• Solution: Lens with finite aperture
image plane
lens
W
scene point
focal point
I
f
v
u
Lens Law:
1
1
u
v
20
1
f